A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(9t^{2},3\right)$ for time $t\geq 0$. At $t=2$, the particle is at the point $(6,-7)$. What is the particle's position at $t=7$ ? $($
Answer: To find the particle's position at $t=7$, we need to find its horizontal displacement $\Delta x$ and its vertical displacement $\Delta y$, and add those to its initial position $(6,-7)$ : $\text{Position at }t=7\text{: }(6+\Delta x,-7+\Delta y)$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between $t=2$ and $t=7$ : $\Delta x=\int_{2}^{7} 9t^{2}\,dt=1005$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between $t=2$ and $t=7$ : $\Delta y=\int_{2}^{7} 3\,dt=15$ Now we can find the particle's position: $\begin{aligned} &\phantom{=}(6+\Delta x,-7+\Delta y) \\\\ &=\left(6+1005,-7+15\right) \\\\ &=(1011,8) \end{aligned}$ In conclusion, particle's position at $t=7$ is $(1011,8)$.